Prepare 10.00 mL of 0.010 M NaOH by diluting the NaOH solution used in Trial 1. (Your procedure should clearly show your calcula
1 answer:
Answer:
<em> = 0.2 mL.</em>
Explanation:
Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:
where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.
Let C1 = 0.5M, V2 = ?
C2 = 0.010M; V2 = 10mL
⇒Volume of stock solution to be diluted, V2
= × 0.010
<em> = 0.2 mL.</em>
Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.
I hope this was a detailed explanation given the missing details of "Trial 1" in the question.
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If the concentration of acetyl chloride is increased ten times the rate of reaction is increased ten times.
The conversion of acetyl chloride to methyl acetate is a substitution reaction. Recall that a substitution reaction is one in which a moiety in a molecule is replaced by another.
In this reaction, the CH3O- ion replaces the chloride ion. In the first step, the CH3O- ion attacks the substrate in a slow step. This creates a tetrahedral intermediate. Loss of the chloride ion yields the methyl acetate product.
The rate determining step is the formation of the tetrahedral intermediate. Since the reaction is first order in the acetyl chloride, if its concentration is increased ten times the rate of reaction is increased ten times.
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